Friday, January 11, 2019

Diophantine Equations Revisited

A Diophantine equation "is an algebraic equation in one or more unknowns, with integer coefficients, for which integer solutions are sought", according to the authors of 'What is Mathematics?'. They add that such an equation may "have no solutions, a finite number of solutions or an infinite number." As the simplest example of such an equation they give: ax + by = c.
Here, a, b and c denote integers and the equation needs to be solved for x and y, and ideally these solutions are integers, not fractions. At this point, we have to cut to the Euclidean algorithm and how it's used. A general theorem may be stated thus: If a is any integer, and b is any integer greater than zero, then we can always find an integer q such that: a = b*q + r
where r is an integer satisfying the inequality: 0 < r < bA general task is to find the “greatest common divisor” (gcd) of a and b, called d(a,b). If d = (a,b)then positive or negative integers k and m can always be found such that: d = ka + mb
This proceeds by successive divisions with remainders. As an example, consider the Euclidean algorithm for finding (a,b) = (61, 24). Here gcd or greatest common divisor is 1 and we find:
61 = 2*24 + 13
24 = 1*13 + 11
13 = 1*11 + 2
11 = 5*2 + 1
2 = 2*1 + 0From the 1st equation we have: 13 = 61 - 2*24
From the end of the second result we get:
11 = 24 – 13 = 24 - 61 - 2*24 = -61 + 3*24
From the third:
2 = 13 – 11 = (61 – 2*24) – (-61 + 3*24) = 2*61 – 5*24

From the fourth:
1 = 11 – 5*2 = (-61 + 3*24) – 5(2*61 – 5*24) = -11.61 + 28*2Apart from these operations, we can also use the Euclidean algorithm to compute continued fractions. The latter are especially useful for solving Diophantine equations. Continued fractions, for their part, are key aspects of a branch of higher arithmetic known as 'Diophantine analysis'. The graphic below illustrates how continued fractions are computed.As another example in more detail, we look at 169/70 and the continued fraction development:

169/70 = 2 + 29/70 = 2 + 1/70/29

70/29 = 2 + 12/29 = 2 + 1/ (29/12)

29/12 = 2 + 5/18 = 2 + 1/ (12/5)

12/5 = 2 + 2/5 = 2 + 1/(5/2)

And: 5/2 = 2 ½

Let's now see how the preceding operations can be used to solve a Diophantine equation. In particular, we want to solve: 7x + 11y = 13 We proceed by working with 7 and 11, respectively, using the Euclidean algorithm.

Thus: 11 = 1*7 + 4 and 7 = 1*4 + 3

We see remainder 4 for the first, so use the algorithm again, so: 4 = 1*3 + 1

In summary then, the linear Diophantine equation of form ax + by = c has a solution in integers if and only if c is a multiple of (a, b). In this case, a particular solution may be found using the Euclidean algorithm.

About Me

Specialized in space physics and solar physics, developed first astronomy curriculum for Caribbean secondary schools, has written thirteen books - the most recent:Fundamentals of Solar Physics. Also: Modern Physics: Notes, Problems and Solutions;:'Beyond Atheism, Beyond God', Astronomy & Astrophysics: Notes, Problems and Solutions', 'Physics Notes for Advanced Level&#39, Mathematical Excursions in Brane Space, Selected Analyses in Solar Flare Plasma Dynamics; and 'A History of Caribbean Secondary School Astronomy'. It details the background to my development and implementation of the first ever astronomy curriculum for secondary schools in the Caribbean.